Question

5. -/1 points DevoreStat9 5.E.041. My Notes Ask Your Teacher Let X be the number of packages being mailed by a randomly selected customer at a certain shipping facility. Suppose the distribution of X is as follows. 1 2 3 4 p(x) 0.2 0.4 0.1 0.3 (a) Consider a random sample of size n = 2 (two customers), and let be the sample mean number of packages shipped. Obtain the probability distribution of x 1 1.5 2 2.5 3 3.5 4 (b) Refer to part (a) and calculate P(X s 2.5). (c) Again consider a random sample of size n = 2, but now focus on the statistic R = the sample range (difference between the largest and smallest values in the sample). Obtain the distribution of R. (Hint: Calculate the value of R for each outcome and use the probabilities from part (a).] PCR) CCCC (d) If a random sample of size n = 4 is selected, what is pls 1.5)? [Hint: You should not have to list all possible outcomes, only those for which is 1.5.) Need Help? Read It Watch Talk to a Tutor

Answer 1

(a)

Following table shows the all possible sample of size n=2:

X1	X2	P(X1=x1)	P(X2=x2)	xbar	P(xbar) = P(X1=x1)*P(X2=x2)
1	1	0.2	0.2	1	0.04
1	2	0.2	0.4	1.5	0.08
1	3	0.2	0.1	2	0.02
1	4	0.2	0.3	2.5	0.06
2	1	0.4	0.2	1.5	0.08
2	2	0.4	0.4	2	0.16
2	3	0.4	0.1	2.5	0.04
2	4	0.4	0.3	3	0.12
3	1	0.1	0.2	2	0.02
3	2	0.1	0.4	2.5	0.04
3	3	0.1	0.1	3	0.01
3	4	0.1	0.3	3.5	0.03
4	1	0.3	0.2	2.5	0.06
4	2	0.3	0.4	3	0.12
4	3	0.3	0.1	3.5	0.03
4	4	0.3	0.3	4	0.09
Total					1

Now we need to add the probabilities of samples for which xbar are same. The pdf of xbar will be as follows:

xbar	P(xabr)
1	0.04
1.5	0.16
2	0.2
2.5	0.2
3	0.25
3.5	0.06
4	0.09

(b)

$P(\bar{x}\leq 2.5)=P(\bar{x}=1)+P(\bar{x}=1.5)+P(\bar{x}=2)+P(\bar{x}=2.5)=0.04+0.16+0.2+0.2=0.6$

(c)

Following tables shows the range of distribution and corresponding probabilities:

X1	X2	P(X1=x1)	P(X2=x2)	Range, R	P(R=r) = P(X1=x1)*P(X2=x2)
1	1	0.2	0.2	0	0.04
1	2	0.2	0.4	1	0.08
1	3	0.2	0.1	2	0.02
1	4	0.2	0.3	3	0.06
2	1	0.4	0.2	1	0.08
2	2	0.4	0.4	0	0.16
2	3	0.4	0.1	1	0.04
2	4	0.4	0.3	2	0.12
3	1	0.1	0.2	2	0.02
3	2	0.1	0.4	1	0.04
3	3	0.1	0.1	0	0.01
3	4	0.1	0.3	1	0.03
4	1	0.3	0.2	3	0.06
4	2	0.3	0.4	2	0.12
4	3	0.3	0.1	1	0.03
4	4	0.3	0.3	0	0.09
Total					1

Following table shows the pdf of range:

Range, R	P(R=r)
0	0.3
1	0.3
2	0.28
3	0.12

(d)

We need to find the samples such that sample sum is less than equal to 6. Following table shows all the possible sample such that sum is less than equal to 6:

X1	X2	X3	X4	xbar	P(X1=x1)	P(X2=x2)	P(X3=x3)	P(X4=x4)	P(xbar)
1	1	1	1	1	0.2	0.2	0.2	0.2	0.0016
1	1	1	2	1.25	0.2	0.2	0.2	0.4	0.0032
2	1	1	1	1.25	0.4	0.2	0.2	0.2	0.0032
1	1	2	1	1.25	0.2	0.2	0.4	0.2	0.0032
1	2	1	1	1.25	0.2	0.4	0.2	0.2	0.0032
1	1	2	2	1.5	0.2	0.2	0.4	0.4	0.0064
2	2	1	1	1.5	0.4	0.4	0.2	0.2	0.0064
1	2	2	1	1.5	0.2	0.4	0.4	0.2	0.0064
1	2	1	2	1.5	0.2	0.4	0.2	0.4	0.0064
2	1	2	1	1.5	0.4	0.2	0.4	0.2	0.0064
2	1	1	2	1.5	0.4	0.2	0.2	0.4	0.0064
1	1	1	3	1.5	0.2	0.2	0.2	0.1	0.0008
1	1	3	1	1.5	0.2	0.2	0.1	0.2	0.0008
1	3	1	1	1.5	0.2	0.1	0.2	0.2	0.0008
3	1	1	1	1.5	0.1	0.2	0.2	0.2	0.0008
Total									0.056

So,

$P(\bar{x}\leq 1.5)=0.056$